# How to solve this first order nonlinear PDE?

Given $u=x{{u}_{x}}+y{{u}_{y}}+\frac{1}{2}\left( u_{x}^{2}+u_{y}^{2} \right)$ , find a solution with $u\left( x,0 \right)=\frac{1}{2}\left( 1-{{x}^{2}} \right)$ .

$\frac{dp}{0}=\frac{dq}{0}=\frac{du}{xp+{{p}^{2}}+qy+{{q}^{2}}}=\frac{dx}{x+p}=\frac{dy}{y+q}.$

$\Rightarrow p=a,q=b\Rightarrow u=ax+by+\frac{1}{2}\left( {{a}^{2}}+{{b}^{2}} \right)$

$u\left( x,0 \right)=\frac{1}{2}\left( 1-{{x}^{2}} \right)=ax+0+\frac{1}{2}\left( {{a}^{2}}+{{b}^{2}} \right).$

$\Rightarrow {{b}^{2}}=\frac{1}{2}\left( 1-{{x}^{2}}-{{a}^{2}} \right)-ax.$

$\Rightarrow u=\frac{1}{2}\left( 1-{{x}^{2}} \right)+y\sqrt{\frac{1}{2}\left( 1-{{x}^{2}}-{{a}^{2}} \right)-ax}\text{ where }a\text{ is an arbitary constant}.$

Solution alternative (for Clairaut's form) :

$dz=pdx+qdy$

$z=ax+by+c\text{ }.$

$u\left( x,0 \right)=\frac{1}{2}\left( 1-{{x}^{2}} \right)=ax+0+c\text{ }\Rightarrow \text{ }c=\frac{1}{2}\left( 1-{{x}^{2}} \right)-ax\text{ }.$

$\Rightarrow u=\frac{1}{2}\left( 1-{{x}^{2}} \right)+by.$

• No,it's alright.Actually ,this is how you derive Clairaut's form conditions. – Koro Nov 14 '15 at 3:41

$u=ax+by+\frac{1}{2}(a^2+b^2)$ then use the initial condition to eliminate $b$